Solve {lll}{4}&{150}&{2}{1}&{125}&{3}{6}&{175}&{3}

$$\left| \begin{array} { l l l } { 4 } & { 150 } & { 2 } \\ { 1 } & { 125 } & { 3 } \\ { 6 } & { 175 } & { 3 } \end{array} \right|$$

Evaluate

$$500$$

Steps Using Sarrus' Rule

Find the determinant of the matrix using the method of diagonals.

$$det(\left(\begin{matrix}4&150&2\\1&125&3\\6&175&3\end{matrix}\right))$$

Extend the original matrix by repeating the first two columns as the fourth and fifth columns.

$$\left(\begin{matrix}4&150&2&4&150\\1&125&3&1&125\\6&175&3&6&175\end{matrix}\right)$$

Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.

$$4\times 125\times 3+150\times 3\times 6+2\times 175=4550$$

Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.

$$6\times 125\times 2+175\times 3\times 4+3\times 150=4050$$

Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

$$4550-4050$$

Subtract 4050 from 4550.

$$500$$

Steps Using Expansion by Minors

Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).

$$det(\left(\begin{matrix}4&150&2\\1&125&3\\6&175&3\end{matrix}\right))$$

To expand by minors, multiply each element of the first row by its minor, which is the determinant of the 2\times 2 matrix created by deleting the row and column containing that element, then multiply by the element's position sign.

$$4det(\left(\begin{matrix}125&3\\175&3\end{matrix}\right))-150det(\left(\begin{matrix}1&3\\6&3\end{matrix}\right))+2det(\left(\begin{matrix}1&125\\6&175\end{matrix}\right))$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the determinant is ad-bc.

$$4\left(125\times 3-175\times 3\right)-150\left(3-6\times 3\right)+2\left(175-6\times 125\right)$$

Simplify.

$$4\left(-150\right)-150\left(-15\right)+2\left(-575\right)$$

Add the terms to obtain the final result.

$$500$$

Factor

$$2^{2}\times 5^{3}$$